Least Squares using the SVD¶

In [1]:
#keep
import numpy as np
import numpy.linalg as la
import scipy.linalg as spla
%matplotlib inline

In [2]:
#keep
# tall and skinny w/nullspace
np.random.seed(12)
A = np.random.randn(6, 4)
b = np.random.randn(6)
A[3] = A[4] + A[5]
A[1] = A[5] + A[1]
A[2] = A[3] + A[1]
A[0] = A[3] + A[1]


Part I: Singular least squares using QR¶

Let's see how successfully we can solve the least squares problem when the matrix has a nullspace using QR:

In [3]:
#keep
Q, R = la.qr(A)

In [4]:
#keep
R.round(3)

Out[4]:
array([[-4.526,  3.492, -0.204, -3.647],
[ 0.   ,  0.796,  0.034,  0.603],
[ 0.   ,  0.   , -1.459,  0.674],
[ 0.   ,  0.   ,  0.   ,  0.   ]])

We can choose x_qr[3] as we please:

In [5]:
#keep
x_qr = np.zeros(A.shape[1])

In [6]:
x_qr[3] = 0

In [7]:
#keep
QTbnew = Q.T.dot(b)[:3,] - R[:3, 3] * x_qr[3]
x_qr[:3] = spla.solve_triangular(R[:3,:3], QTbnew, lower=False)


Let's take a look at the residual norm and the norm of x_qr:

In [8]:
#keep
R.dot(x_qr)-Q.T.dot(b)[:4]

Out[8]:
array([ -4.44089210e-16,   0.00000000e+00,   0.00000000e+00,
-1.97736227e-01])
In [9]:
#keep
la.norm(A.dot(x_qr)-b, 2)

Out[9]:
2.1267152888030982
In [10]:
#keep
la.norm(x_qr, 2)

Out[10]:
0.82393512974131566

Choose a different x_qr[3] and compare residual and norm of x_qr.

Part II: Solving least squares using the SVD¶

Now compute the SVD of $A$:

In [11]:
U, sigma, VT = la.svd(A)


Make a matrix Sigma of the correct size:

In [12]:
#keep
Sigma = np.zeros(A.shape)
Sigma[:4,:4] = np.diag(sigma)


And check that we've actually factorized A:

In [13]:
#keep
(U.dot(Sigma).dot(VT) - A).round(4)

Out[13]:
array([[ 0., -0.,  0.,  0.],
[ 0., -0.,  0.,  0.],
[ 0., -0.,  0.,  0.],
[ 0., -0., -0.,  0.],
[ 0., -0.,  0.,  0.],
[ 0., -0., -0.,  0.]])

Now define Sigma_pinv as the "pseudo-"inverse of Sigma, where "pseudo" means "don't divide by zero":

In [14]:
Sigma_pinv = np.zeros(A.shape).T
Sigma_pinv[:3,:3] = np.diag(1/sigma[:3])
Sigma_pinv.round(3)

Out[14]:
array([[ 0.147,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ],
[ 0.   ,  0.624,  0.   ,  0.   ,  0.   ,  0.   ],
[ 0.   ,  0.   ,  1.055,  0.   ,  0.   ,  0.   ],
[ 0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  0.   ]])

Now compute the SVD-based solution for the least-squares problem:

In [15]:
x_svd = VT.T.dot(Sigma_pinv).dot(U.T).dot(b)

In [16]:
#keep
la.norm(A.dot(x_svd)-b, 2)

Out[16]:
2.1267152888030978
In [17]:
la.norm(x_svd)

Out[17]:
0.77354943014895816
• What do you observe about $\|\text{x_svd}\|_2$ compared to $\|\text{x_qr}\|_2$?
• Is $\|\text{x_svd}\|_2$ compared to $\|\text{x_qr}\|_2$?
In [ ]: